Problem 4

Let ABC be an acute triangle. Let D be the foot of the perpendicular from A to BC, M the midpoint of BC, and H the orthocenter of ABC. Let E be the intersection of the circumcircle Gamma of ABC with the ray MH, and let F be the other intersection of line ED with Gamma. Prove that BF/CF = AB/AC, where XY denotes the length of segment XY.

Step 6 of 6: Finish with the midpoint

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BM=CM\Longrightarrow\frac{FC}{AC}=\frac{FB}{AB}\Longrightarrow\boxed{\frac{BF}{CF}=\frac{AB}{AC}}.

Detailed analysis

Since M is the midpoint of BC, BM=CM. Equating the two expressions from the similarities and rearranging gives BF/CF=AB/AC, as required.